Anomalies in Global Symmetries (Srednicki ch 76) In chapter 76 of Srednicki's QFT book, he defines $C^{\mu\nu\rho}(p,q,r)$ via (76.21)
\begin{equation}
(2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)\equiv \int d^{4}xd^{4}yd^{4}z e^{-i(px+qy+rz)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
and rewrites (76.15)
\begin{equation}
\langle p,q|j^{\rho}_{\textrm{A}}(z)|0\rangle = (ig)^{2}\epsilon_{\mu}\epsilon'_{\nu} \int d^{4}xd^{4}y e^{-i(px+qy)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
as (76.22)
\begin{equation}
\langle p,q|j^{\rho}_{\textrm{A}}(z)|0\rangle = -g^{2}\epsilon_{\mu}\epsilon'_{\nu} C^{\mu\nu\rho}(p,q,r)e^{irz}|_{r=-p-q}
\end{equation}
but I don't understand this transformation.
When I multiply (76.21) by $e^{irz}|_{r=-p-q}$ on both sides, I get
\begin{equation}
(2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)e^{irz}|_{r=-p-q}= \int d^{4}xd^{4}yd^{4}z e^{-i(px+qy)}e^{-i(p+q+r)z}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
It seems though that the right hand side cannot be written  as 
\begin{equation}
(2\pi)^{4}\delta^{4}(p+q+r) \int d^{4}xd^{4}y e^{-i(px+qy)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle
\end{equation}
because there is a $z$ dependence in $j^{\rho}_{\textrm{A}}(z)$ and thus (76.22) doesn't hold. Am I wrong?
 A: Prof. Srednicki multiplies the first line that you wrote,
$(2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)\equiv \int d^{4}xd^{4}yd^{4}w e^{-i(px+qy+rw)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(w)|0\rangle$,
by $e^{i r z}$ and integrates on $r$.
This is explicitly $$\int d^4r e^{i r z} (2\pi)^{4}\delta^{4}(p+q+r)C^{\mu\nu\rho}(p,q,r)= \int d^4r \int d^{4}xd^{4}yd^{4}w e^{-i(px+qy)}e^{ir(w-z)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(w)|0\rangle$$
The left side is easily evaluated with the delta function. The right side, if one integrates on $r$ first, enjoys a factor of $(2\pi)^4 \delta(w-z)$.
Thus we have
$(2\pi)^{4}e^{i r z}C^{\mu\nu\rho}(p,q,r)|_{p+q+r=0} = (2\pi)^{4}\int d^{4}xd^{4}yd^{4}w e^{-i(px+qy)}\delta(w-z)\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(w)|0\rangle .$
Dividing both sides by $(2 \pi)^4$ and integrating over $w$ nets
$e^{i r z}C^{\mu\nu\rho}(p,q,r)|_{p+q+r=0} = \int d^{4}xd^{4}y e^{-i(px+qy)}\langle 0|\textrm{T}j^{\mu}(x)j^{\nu}(y)j^{\rho}_{\textrm{A}}(z)|0\rangle .$
Using your second line gives us the desired result.
